Chebyshev polynomials, quadratic surds and a variation of Pascal s triangle Roland Bacher To cite this version: Roland Bacher. Chebyshev polynomials, quadratic surds and a variation of Pascal s triangle. IF_PREPUB. 015. <hal-0106138> HAL Id: hal-0106138 https://hal.archives-ouvertes.fr/hal-0106138 Submitted on 8 Sep 015 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.
Chebyshev polynomials, quadratic surds and a variation of Pascal s triangle Roland Bacher September 8, 015 Abstract 1 : Using iterated Chebyshev polynomials of both inds, we construct rational fractions which are convergents of the smallest root of x αx+1 for α 3,4,5,... Some of the underlying identities suggest an identity involving binomial coefficients which leads to a triangular array sharing many properties with Pascal s triangle. 1 Introduction refs: Watson-Whittacer, qq chose sur fractions continues, polys de Cheb. Chebyshev polynomials of the firstind T 0,T 1,... and of the second ind U 0,U 1,... have recursive definitions given by and by T 0 x) 1,T 1 x) x,t n+1 x) xt n x) T n 1 x),n 1, U 0 x) 1,U 1 x) x,u n+1 x) xu n x) U n 1 x),n 1. We write in the sequel always simply T n,u n for T n x),u n x). The polynomial sequences T 0,T 1,... and U 0,U 1,... satisfy the same linear recursion relation with characteristic polynomial Z xz + 1. This implies easily the formulae x+ ) n x 1 + x ) n x 1 T n 1) This wor has been partially supported by the LabEx PERSYVAL-Lab ANR 11- LABX-005). The author is a member of the project-team GALOIS supported by this LabEx. 1 Keywords: Chebyshev polynomial, continued fraction, binomial coefficient. Math. class: Primary: 33C45, Secondary: 11A55, 11B65 1
and U n x+ ) n+1 x 1 x ) n+1 x 1 x 1. ) The identity cosxcosnx cosn+1)x+cosn 1)x together with the initial values T 0 1,T 1 x) impliesby aneasy inductiont n cosx) cosnx which is often used for defining Chebyshev polynomials of the first ind). As a consequence, we have the identity T nm T n T m. 3) For n N and d 1 we introduce the rational fraction S n,d S n,d x) n 1 U d 1 T d+1). 4) U d T d+1) j 0 j0 Theorem 1.1 We have for all n N and for all d 1 the identity S n,d xs n,d +1 n 1 U d T d+1) j j0. 5) Corollary 1. For x evaluated to a real number in R \ [ 1,1], the limit lim n+d S n,d exists and is given by the root signx) x ) x 1 where signx) {±1} denotes the sign of x) closest to zero of X xx +1. The following result expresses S n,d as a simple fraction: Theorem 1.3 We have S n,d U d+1) n+1. U d+1) n+1 1 Corollary 1. is now an almost immediate consequence of Formula ) and Theorem 1.3. Note that Formula 4), perhaps computed using iteratively 3), is perhaps better suited for computations than the simpler expression given by Theorem 1.3. Finally we have the following result: Theorem 1.4 We have U n U n+1 [0;x 1,1,x 1),1) n ].
Theorem 1.3 and Theorem 1.4 together imply that the evaluation at an integer x of S n,d is a convergent of the real quadratic surd x+ x 1 [0;x 1,1,x 1),1,x 1),1,x 1)] with minimal polynomial X xx +1 Z[X]. In a last only losely related part we study some identities involving binomial coefficients obtained by expressing Chebyshev polynomials in terms of binomials coefficients and generalizing some of the previously obtained identities). They lead to the array of numbers 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 4 3 1 1 0 3 6 7 4 1 1 1 3 9 13 11 5 1 1 0 4 1 4 16 6 1 mimicing several aspects of Pascal s triangle. The sequel of the paper is organised as follows: Section describes and proves useful identities among Chebyshev polynomials. Section 3 contains a proof of Theorem 1.1, 1.3 and 1.4. A final Section describes a few relations with binomial coefficients and studies a few features of the above analogue of Pascal s triangle. Analogues of Vajda s identity for Chebyshev polynomials Fibonacci numbers F 1 F 1,F n+1 F n +F n 1 satisfy Vajda s identity F n+i F n+j F n F n+i+j 1) n F i F j. The following result describes analogues for Chebyshev polynomials: Theorem.1 We have the following identities for Chebyshev polynomials: U n+i U n+j U n 1 U n+1+i+j U i U j T n+i T n+j T n 1 T n+1+i+j 1 x )U i U j T n+i U n+j U n 1 T n+1+i+j T i U j T n+i U n+j T n 1 U n+1+i+j U i T j+ T n+i T n+j x 1)U n 1 U n 1+i+j T i T j 3
The case i j 0 of the first identity specialises to the so-called Cassini-Simpson identity U n U n+1 U n 1 1 and implies Turan s inequality U nx) > U n+1 x)u n 1 x) for all real x. The last equation generalises the instance T n x 1)U n 1 1 corresponding to i j 0) of Pell s equation. Only a few cases of the first identity will in fact be used in the sequel. Proof of Theorem.1 We consider Rn,i,j) U n+i U n+j U n 1 U n+1+i+j U i U j. We have to show that Rn,i,j) 0 for all n 1 and for all i,j N. Using the recursion relation U n+1 xu n U n 1 on all terms depending on i, respectively depending on j, we see that it is enough to prove the equalities Rn,i,j) 0 for i,j {0,1}. Using the obvious identity Rn,i,j) Rn,j,i) we are left with three cases: Rn,0,0),Rn,0,1) and Rn,1,1). The computation U n U n 1 U n+1 U 0 U n xu n 1 U n ) U n 1 xu n U n 1 ) 1 U n 1 U nu n 1 shows Rn,0,0) Rn 1,0,0) for n. Similarly, U n+1 U n U n 1 U n+ U 1 U 0 xu n U n 1 )U n U n 1 xu n+1 U n ) x xu n U n 1U n+1 1) shows Rn,1,0) Rn,0,1) xrn,0,0). Finally, the identities U n+1 U n+1 U n 1 U n+3 U 1 U n+1 xu n U n 1 ) U n 1 xu n+ U n+1 ) 4x xu n+1 U n U n 1 U n+ x) show Rn,1,1) xrn,1,0). It is now enough to chec that Rn,i,j) 0 for n {1,} and i,j {0,1}. Proofs of the remaining identities are similar. Remar. Short direct proofs of Theorem.1 can be obtained using Formulae 1) and ). 4
3 Proof of Theorem 1.1, 1.3 and 1.4 3.1 Useful identities The following result is well-nown: Lemma 3.1 We have for all d 1 the identity xu d U d 1 +T d+1. The proof is an easy induction left to the reader. Lemma 3. We have for all d 1 the identity U d U d 1 +T d+1u d 1 +1. Proof Using Lemma 3.1 and the recursive definition of U, Lemma 3. is equivalent to U d U d 1 +xu d U d 1 )U d 1 +1 U d 1 +xu d)u d 1 +1 U d 1 +U d+1 +U d 1 )U d 1 +1 U d+1 U d 1 +1 which is a special case of the first equality in Theorem.1. Lemma 3.3 We have T d U n U n+d +U n d for all n N and for all d {0,1,...,n}. Proof We set Rn,d) T d U n U n+d U n d. Since T 0 1 and T 1 x we have Rn,0) Rn,1) 0. The identities Rn,d) T d U n U n+d U n d xt d 1 T d )U n xu n+d 1 U n+d ) xu n d+1 U n d+ ) xrn,d 1) Rn,d ) finish the proof. Lemma 3.4 We have for all n 1 the identity U n 1 T n U n 1. 5
Proof Equality holds for n 1. Using Lemma 3.1) we have for n the identities U n 1 T n U n 1 U n 1 xu n 1 U n )U n 1 U n 1 xu n 1 U n )U n 1 +U n 1 U n U n U n 1 +U n 1 +U n 1 U n U 1+n 1) U 1+n ) U 1 1 U 1+1+n 1)+n ) U n 1 U n ). The last expression equals zero by the first identity of Theorem.1. Lemma 3.5 We have for all n and for all d 1. U n 1)d 1 U n 1 T d U nd 1 U n T d Proof The case n boils down U d 1 U d 1 T d which holds by Lemma 3.4). Adding to 0 T d U n 1)d 1 U n )d 1 U nd 1 )U n T d which holds by Lemma 3.3) the induction hypothesis we get 0 T d U n 1)d 1 U n )d 1 U nd 1 )U n T d +U n )d 1 U n T d U n 1)d 1 U n 3 T d U n 1)d 1 T d U n T d U n 3 T d ) U nd 1 U n T d U n 1)d 1 U n 1 T d U nd 1 U n T d which ends the proof. 3. Proof of Theorem 1.1 We prove first that equation 5) holds for n 0. Multiplying the left-side of equation 5) by Ud, we get U d 1 xu d 1U d +U d U d 1 U d 1U d+1 +U d 1 )+U d U d U d 1U d+1 U 0 1 by applying the recursive definition of U i and the first identity of Theorem.1 with n d,i j 0. 6
Setting x T d+1 in equation 5) and dividing the result by Ud, we have now by induction Sn,d T d+1 U d ) S n,d T d+1 T d+1 Ud + 1 Ud n+1 1 U d T d+1) j where we have also used 3) on the right side. We rewrite now the obvious identity as j0 6) S n+1,d U d 1 +S n,d T d+1 U d. 7) S n,d T d+1 U d S n+1,d U d 1. 8) Using 8) the left side of 6) equals ) Ud S n+1,d U d 1 U d S n+1,d U d 1 T d+1 U d Ud + 1 Ud Sn+1,d U d 1 S n+1,d + U d 1 U d Ud T d+1 U d 1 S n+1,d +T d+1 U d Ud + 1 Ud Since U d 1 S n+1,d T d+1 S n+1,d xs n+1,d U d U d by Lemma 3.1) and U d 1 U d +T d+1 U d 1 U d + 1 U d by Lemma 3.), we get finally Sn+1,d xs n+1,d+1 for the left side of 6). This ends the proof. 3.3 Proof of Theorem 1.3 Equality holds obviously for n 0. Applying the induction hypothesis to 7) we have to establish the equality 1 equivalent to U d U d+1) n+ U d+1) n+ 1 U d 1 + U d+1) n+1 T d+1 U d+1) n+1 1 T d+1 0 U d U d+1) n+ U d 1 U d+1) n+ 1)U d+1) n+1 1 T d+1 U d+1) n+ 1U d+1) n+1 T d+1 U d+1) n+ du d+1) n+1 1 T d+1 U d+1) n+ 1U d+1) n+1 T d+1 7
where we have applied the first identity of Theorem.1 with n d,i 0,j d+1) n+ d. The identity 0 U d+1) n+ du d+1) n+1 1 T d+1 U d+1) n+ 1U d+1) n+1 T d+1 is now the case n,d) d 1) n+1,d+1) of Lemma 3.5. 3.4 Proof of Theorem 1.4 Proof of Theorem 1.4 Equality holds for n 0. Setting γ n [0;x 1,1,x 1),1) n ] we have γ n [0;x,1,x 1),1) n 1] 1 γ n showing γ n+1 1 1 x 1+ 1+ γn 1 γn 1. x γ n The result follows now by induction from the trivial identities 1 U n+1 U n+1. x Un xu U n+1 U n U n+ n+1 An easy computation shows the continued fraction expansion x x 1 [0;x 1,1,x 1),1,x 1),1,x 1)]. for x {,3,4,...}. Equality follows thus from analytic continuation whenever both sides mae sense. Combining Theorem 1.3 and Theorem 1.4 we see that S n,d [0;x 1,1,x 1),1) ] d+1)n+1 using a hopefully self-explanatory notation) is a convergent of x x 1 for x,3,... 8
4 A sum of products of two binomial coefficients 4.1 Coefficients of Chebyshev polynomials Lemma 4.1 Explicit expressions for coefficients of Chebyshev polynomials are given by the formulae T n 1 n/ ) )) n+1 n 1 1) x) n, U n 0 ) n 1) x) n 0 ) 1, 1, n/ using the conventions 1 ) 0, 1 1 ) 1 ) 1, 0). Proof The formulae hold obviously for T 0,T 1 and U 0,U 1. We have now and T n+1 xt n T n 1 x 1 n/ 0 1) n+1 ) n 1 1 n 1)/ ) n n 1) 0 1 ) n+1 n+1 1) + 1 1 ) n 1 n 1 1) + 3 1 ) n+ n 1) 1 )) x) n )) x) n 1 )) x) n+1 )) x) n+1 )) x) n+1 U n+1 xu n U n 1 n/ ) n x 1) 0 n 1)/ 0 1) n x) n ) n 1 1) ) n + 1 ) n+1 1) x) n+1. These identities imply the result by induction. x) n 1 )) x) n+1 9
4. A curious identity Rewriting Chebyshev polynomials in terms of binomial coefficients using the identities of Lemma 4.1, some identities among Chebyshev polynomials are special cases of the following result. Theorem 4. The expression d n ) ) a+d+x d+ x fa,d,n) x d n 0 is constant in x and depends only on a,d C and n d N {d,d 1,d,d 3,...}. Observethatallvaluesfa,d,n) x aredeterminedbythevaluesf0,d,n) x using the trivial identity fa,d,n) x fa c,d+c,n+c) x+c 9) with c a/. Theorem 4. implies that Q N z) f0,z/,z/ N) is a polynomial in Q[z] of degree N such that Q N Z) Z. Lemma 4.3 We have the identities and fa,d,n) x fa 1,d,n) x +fa 1,d,n+1) x 1 10) fa,d,n) x fa 1,d,n) x+1 +fa 1,d,n+1) x+1 11) Proof Follows from the computations and fa,d,n) x d n ) a 1+d+x + 0 a 1+d+x 1 )) ) d+ x d n fa 1,d,n) x + ) ) a 1+d+x 1 1) d+ 1) x 1) 1 d n+1) 1) fa 1,d,n) x +fa 1,d,n+1) x 1 fa,d,n) x d n a+d+x 0 ) d+ x 1 d n fa 1,d,n) x+1 +fa 1,d,n+1) x+1 10 ) + )) d+ x 1 d n 1
Proof of Theorem 4. Since ) x xx 1) x +1)!, the function fa,d,n) x is a polynomial of degree at most d n in x. It is thus independent of x for n d. Subtracting equation 10) from 11) we get fa 1,d,n) x+1 fa 1,d,n) x fa 1,d,n+1) x 1 fa 1,d,n+1) x+1 which implies the result by induction on d n. 4.3 A few properties of f0,d,n) The numbers l i,j f0,i 1)/,j i+1)/) i j i 1 0 +x ) i 1 + x i j with i N and j {0,...,i} and x arbitrary) form the Pascal-lie triangle: 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 4 3 1 1 0 3 6 7 4 1 1 1 3 9 13 11 5 1 1 0 4 1 4 16 6 1 as shown by the following result: Proposition 4.4 We have l i,0 1) i, l i,i 1 and l i,j l i 1,j 1 +l i 1,j. Proof Using x i 1 we get for j 0 the evaluation f 0, i 1 ), i+1 ) i ) ) i 1 i 0 ) ) 1 i 1) i. i 0 For i j we get l i,i f 0, i 1, i 1 i+1 ) 1 by a trivial computation. 11
Using 10) followed by two applications of 9) with c 1 we have l i,j f 0, i 1 ) i+1,j f 1, i 1 ) i+1,j +f 1, i 1 ) i+1,j +1 f 0, i 1 1 i+1,j 1 ) +f 0, i 1 1 i+1,j + 1 ) f 0, i 1) 1,j 1 i 1)+1 ) +f 0, i 1) 1,j i 1)+1 ) l i 1,j 1 +l i 1,j which proves the result. 4.4 An LU-decomposition Interpreting the integers l i,j,i,j 0 as the coefficients of an infinite unipotent matrix L and introducing similarly the matrix M with coefficients M i,j l i+j,j, we have the following result: Proposition 4.5 We have M LU where U is the upper-triangular matrix with coefficients U i,j j i),i,j 0. In particular, we have detmn)) 1 where Mn) is the square matrix consisting of the first n rows and columns of M. Proposition 4.5 is a special case of the following more general result: Weassociatetwoinfinitematricestoaninfinitesequenceα 0 1,α 1,α,... in a commutative ring with 1 as follows: ThefirstmatrixMα)withcoefficientsM i,j indexedbyi,j Nisdefined recursively by M 0,0 1,M 0,j 1,M i,0 α i,m i,j M i 1,j +M i,j 1,i,j 1. The coefficients M i,j for j > 0 are also given by the formula M i,j i ) +j 1 α i. 0 The second matrix is the unipotent lower-triangular matrix Lα) with lower triangular coefficients L i,j M i j,j,i j 0. It satisfies L i,j L i 1,j + L i 1,j 1 for i,j 1. 1
Proposition 4.6 We have Mα) Lα)U where Mα),Lα) are as above and where U is unipotent upper-triangular with coefficients U j i) given by binomial coefficients. Proof We have obviously M i,j LU) i,j if i 0 or j 0. The remaining cases follow by induction on i+j from the equalities L i, U,j L i, U,j 1 + L i, U 1,j 1 L i, U,j 1 + L i 1, U,j 1 M i,j 1 + M i 1,j 1 M i,j 1 +M i 1,j M i,j where M M1,1 + α 1,1 + α 1 + α,1 + α 1 + α + α 3,...), respectively L L1,1+α 1,1+α 1 +α,...), is obtained from M by removing its first row, respectively from L by removing its first row and column. 4.5 A few more identities The following results show other similarities between l i,j and binomial coefficients: Proposition 4.7 i) We have for all n, N the equality n ) ln, + l n,+1. ii) We have for all n the identity x n 1) n +x+1) n l n, x 1) 1. For proving assertion i), it is enough to chec the equality for all n with 0,1. The general case follows from the last equality in Proposition 4.4. The second assertion holds for n 0. We have now n 1) n +x+1) l n, x 1) 1 1) n +x+1) 1 1 n l n 1, 1 +l n 1, )x 1) 1 1 ) n 1 x 1) 1) n 1 +x+1) l n 1, x 1) 1 13 1
n 1 + 1) n 1 +x+1) l n 1, x 1) 1 1 + 1) n +x+1) 1) n 1 x 1) 1) n 1 1) n 1 x 1)x n 1 +x n 1 x n by induction. 4.6 A few integer sequences A few integer sequences related to the numbers l i,j appear seemingly in [1] proofs are probably easy in most cases). Observe that the array l i,j with its first row removed appears in [1], l 1,1,l,1,l,,l 3,1,l 3,,l 3,3,l 4,1,l 4,,... is A5960 of [1]. The sequence 1,0,,6,,80,96,1106,...,l n,n,n 0 of central coefficients coincides seemingly with A7547 of [1]. It is easy to show that the sequence 1,0,,,6,10,,...,s n n 0 l n, of row-sums is given by s 0 1,s n s n 1 + 1) n ) s n +s n 1. The closely related sequence 1 s n+1 coincides with A1045 of [1]. 1,1,4,8,0,44,100,...,a n n 0 l n, +1) coincides with A8419 of [1]. 0,0,1,3,9,3,57,135,313,711,..., n l n, 1) coincides up to a shift of the index) with A45883 of [1]. 1,,6,14,34,78,178,..., n 0 l n,+1)coincidesseeminglywitha59570 of [1]. There are certainly other sequences of [1] related to the numbers l i,j. Interestingly, the descriptions of the above sequences are lined to several different and apparently unrelated mathematical areas. Acnowledgements I than Bernard Parisse for a useful discussion. References [1] The On-Line Encyclopedia of Integer Sequences, published electronically at http://oeis.org, 015. Roland BACHER, Univ. Grenoble Alpes, Institut Fourier, F-38000 Grenoble, France. e-mail: Roland.Bacher@ujf-grenoble.fr 14